2
$\begingroup$

I have zero inflated response variable I am trying to predict. I am facing few issues applying different regression models that should correct for this.

This is my 10,000 obs dataframe

    e_weight       left_size         right_size        time_diff        
 Min.   :0.000   Min.   :  1.000   Min.   :  1.000   Min.   :      737  
 1st Qu.:0.000   1st Qu.:  1.000   1st Qu.:  1.000   1st Qu.:  4669275  
 Median :0.000   Median :  3.000   Median :  3.000   Median : 12263474  
 Mean   :0.022   Mean   :  6.194   Mean   :  5.469   Mean   : 21000288  
 3rd Qu.:0.000   3rd Qu.:  5.000   3rd Qu.:  5.000   3rd Qu.: 25420278  
 Max.   :3.000   Max.   :792.000   Max.   :792.000   Max.   :155291532

Here the frequency count for my 3 variables enter image description here Indeed I have a problem with zeros...

I tried respectively a Zero-Inflated Negative Binomial Regression and a Zero-inflated Poisson Regression

library(pscl)
m1 <- zeroinfl(e_weight ~ left_size*right_size | time_diff, data = s)
summary(m1)

# Call:
# zeroinfl(formula = e_weight ~ left_size * right_size | time_diff, data = s)
#
# Pearson residuals:
#     Min      1Q  Median      3Q     Max 
# -1.4286 -0.1460 -0.1449 -0.1444 19.6054 
#
# Count model coefficients (poisson with log link):
#                        Estimate Std. Error z value Pr(>|z|)    
#  (Intercept)          -3.8826386  0.0696970 -55.707  < 2e-16 ***
#  left_size             0.0022261  0.0006195   3.594 0.000326 ***
#  right_size            0.0033622         NA      NA       NA    
#  left_size:right_size  0.0001715         NA      NA       NA    
# 
# Zero-inflation model coefficients (binomial with logit link):
#               Estimate Std. Error  z value Pr(>|z|)    
# (Intercept)  1.753e+01  6.011e+00    2.916  0.00354 ** 
#  time_diff   -3.342e-04  1.059e-06 -315.773  < 2e-16 ***
#  ---
#  Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
#
# Number of iterations in BFGS optimization: 28 
#  Log-likelihood: -1053 on 6 Df
#  Warning message:
#  In sqrt(diag(object$vcov)) : NaNs produced

and 

library(MASS)
m2 <- glm.nb(e_weight ~ left_size*right_size + time_diff, data = s)

which gives

There were 22 warnings (use warnings() to see them)
warnings()
Warning messages:
1: glm.fit: algorithm did not converge
...
21: glm.fit: algorithm did not converge
22: In glm.nb(e_weight ~ left_size * right_size + time_diff,  ... :
  alternation limit reached

If I ask a summary for the second model 

summary(m2)

# Call:
# glm.nb(formula = e_weight ~ left_size * right_size + time_diff, 
#     data = s, init.theta = 0.1372733321, link = log)
#
# Deviance Residuals: 
#     Min       1Q   Median       3Q      Max  
# -3.4645  -0.2331  -0.1885  -0.1266   2.7669  
# 
# Coefficients:
#                        Estimate Std. Error z value Pr(>|z|)    
# (Intercept)          -3.239e+00  1.090e-01 -29.699  < 2e-16 ***
# left_size            -4.462e-03  1.835e-03  -2.431 0.015047 *  
# right_size           -7.144e-03  2.118e-03  -3.374 0.000742 ***
# time_diff            -6.013e-08  8.584e-09  -7.005 2.48e-12 ***
# left_size:right_size  4.691e-03  2.749e-04  17.068  < 2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# (Dispersion parameter for Negative Binomial(0.1374) family taken to be 1)
# 
#     Null deviance: 1106.5  on 9999  degrees of freedom
# Residual deviance:  958.5  on 9995  degrees of freedom
# AIC: 1967.2
# 
# Number of Fisher Scoring iterations: 12
# 
# 
#              Theta:  0.1373 
#          Std. Err.:  0.0223 
# Warning while fitting theta: alternation limit reached 
#
#
# 2 x log-likelihood:  -1955.2260

Also both models have very low p-values for heteroskedasticity 

bptest(m1)
# 
#   studentized Breusch-Pagan test
#
# data:  m1
# BP = 244.832, df = 3, p-value < 2.2e-16
#
bptest(m2)
# 
#   studentized Breusch-Pagan test
#
# data:  m2
# BP = 277.2589, df = 4, p-value < 2.2e-16

How should I approach this regression. Would make sense to simply add 1 to all my dataframe before running any regression?

Improve this question
$\endgroup$
1
$\begingroup$

In this case, where your data is simply overwhelmed by the number of zeros, IMO it makes the most sense to split this into a frequency model and a severity model. I.e. something like a logistic regression (or whatever such model you prefer) for whether the data is non-zero, and then a separate model on the severity conditioned on the weight being non-zero.

(Edit: removed my misunderstanding of the problem).

What exactly are you trying to do?

Improve this answer
$\endgroup$
6
  • $\begingroup$ left_size and right_size are never zero. Mostly they take 1 as value (see summary). I am modelling a network, I want to predict the value e_weight of the edge between endpoints left and right based on the time difference (time_diff) between the two nodes and the interaction between the size of the two nodes (left_size and right_size) $\endgroup$
    – Francesco
    Mar 18 '14 at 22:14
  • $\begingroup$ I guess the Zero-Inflated Negative Binomial Regression is doing exactly that: splitting the model into two parts (zeroinfl(e_weight ~ left_size*right_size | time_diff, data = s)): A logit model on time_diff and a count model on left_size*right_size still I guess I don't even enough non zero response values... $\endgroup$
    – Francesco
    Mar 18 '14 at 22:19
  • $\begingroup$ Gotcha, my mistake about the variable values. The key difference with the ZINB model, at least the one that you've specified, is that you've got different regressors for the zero component(time_diff) and for the severity component (left*right). I'm going to edit my original answer... $\endgroup$
    – Mike Nute
    Mar 18 '14 at 22:48
  • $\begingroup$ So it doesn't make sense to have different regressors in the two parts? $\endgroup$
    – Francesco
    Mar 18 '14 at 22:53
  • $\begingroup$ Well it certainly could; that would imply that you believe time_diff impacts existence of weight and left/right impacts size. The more I look at this, the more that first model you've run seems like a legitimate model. If it were me, I'd run it separately as two different models just to keep myself organized. The zero-mass in the Poisson-half of the model above won't have a huge impact if it's allocated to the logistic component. And since most of the variance is in the frequency component, those parameters would be more useful. The severity model may give you some residual signal also. $\endgroup$
    – Mike Nute
    Mar 18 '14 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.